What does this notation mean? Linear map A = [A^\mu \nu]_\mu \nu

In summary, the conversation discusses the decomposition of a vector space V into smaller vector spaces, and the representation of a linear map A in terms of this decomposition. The decomposition allows for writing A in a matrix form, with each block representing a linear map from one smaller vector space to another. This rewriting is possible for any linear map A, without using any special properties of A. The notation \mathcal V^\mu refers to different vector spaces labeled by mu.
  • #1
nonequilibrium
1,439
2
It's used in a certain proof that I'm reading. A is a linear map from a vectorspace V onto itself.

They say they can rewrite the vector space as [itex]\mathcal V = \bigoplus_\mu \mathbb C^{m_\mu} \otimes \mathcal V^\mu[/itex] and I understand this, but they then claim one can (always, as any linear map) rewrite A as [itex]A = [A^{\mu \nu}]_{\mu \nu}[/itex] "where [itex]A^{\mu \nu}[/itex] is a linear map of [itex]\mathbb C^{m_\nu} \otimes \mathcal V^\nu[/itex] to [itex]\mathbb C^{m_\mu} \otimes \mathcal V^\mu[/itex]."

I don't understand the nature of this decomposition/rewriting. Note that this rewriting has to be possible for any A, it doesn't use any special properties of A (that comes later in the proof).
 
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  • #2
This seems to be an extremely obtuse way of writing the matrix of A in a given basis (indexed by \mu in the domain and by \nu in the range).
 
  • #3
Writing a matrix in a given basis is possible, but they also fix the codomain for each part of the domain (it seems); I don't see why that is possible.
 
  • #4
Do they explicitly state what they mean by [itex]\mathcal V^\mu[/itex]?
 
  • #5
Simply different vector spaces labelled by mu, it's as general as that.
 
  • #6
Okay, so they are just writing A in terms of a given basis.

For simplicity, let's suppose [itex]V = V^1 \oplus V^2[/itex]. Then, with respect to this decomposition, we can write
[tex]A = \begin{pmatrix} A^{11} & A^{12} \\ A^{21} & A^{22}\end{pmatrix}.[/tex]
Here A^{11} will take V^1 to V^1, and A^{12} will take V^2 to V^1, etc.
 
  • #7
Oh that makes sense, thank you :)
 

Related to What does this notation mean? Linear map A = [A^\mu \nu]_\mu \nu

1. What is a linear map?

A linear map is a mathematical function that preserves the structure of vector spaces. In other words, it maps one vector space to another in a way that preserves operations such as addition and scalar multiplication.

2. What does the notation [A^\mu \nu]_\mu \nu represent?

This notation represents the components of a linear map A. The first subscript, \mu, represents the row number and the second subscript, \nu, represents the column number. So, [A^\mu \nu]_\mu \nu refers to the component of A in the \mu-th row and \nu-th column.

3. What do the superscripts in the notation represent?

The superscripts represent the basis vectors of the vector spaces involved in the linear map. In this case, A^\mu \nu represents the \mu-th basis vector of the domain space and the \nu-th basis vector of the codomain space.

4. How do I read the notation [A^\mu \nu]_\mu \nu?

The notation is read as "the component of A in the \mu-th row and \nu-th column". It is important to note that the subscripts are often written as subscripts below the main notation, but they can also be written as superscripts on the right side as shown.

5. Can you give an example of a linear map using this notation?

Sure, let's say we have a linear map T: \mathbb{R}^3 \to \mathbb{R}^2, where T is represented by the matrix A = [A^\mu \nu]_\mu \nu = \begin{bmatrix} 3 & 1 & -2 \\ 2 & 0 & 1 \end{bmatrix}. This means that T maps a 3-dimensional vector to a 2-dimensional vector and the \mu-th basis vector of the domain space corresponds to the \mu-th row of the matrix A. So, for example, T(1, 0, 0) = (3, 1) and T(0, 1, 0) = (1, 0).

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